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maxonik [38]
3 years ago
5

This week Emma earned ten more than half the number of dollars she earned last week a babysitting. If this week she earned 100 d

ollars how much did she earn last week
Mathematics
2 answers:
Lynna [10]3 years ago
5 0
180, 180, 180, 180, 180
REY [17]3 years ago
3 0

Answer:180

Step-by-step explanation:

100 - 10 is 90

90 x 2 is 180

I did a reverse calculation you can also check this is correct by knowing that half of 180 is 90 and 100 dollars is 10 more than 90 dollars.

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What is the length of the diagonal?
JulijaS [17]

Answer:

\sqrt{865} feet

Step-by-step explanation:

Using pythagoras theorem,

length \:  =  \sqrt{17 {}^{2}  + 24 {}^{2} }  \\  =  \sqrt{865} ft

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3 years ago
Could the following be points on the graphs of - x + 2y = 5
AlexFokin [52]

Answer:

yeah pretty sure

Step-by-step explanation:

2 is y -1 is x

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3 years ago
Consider an object moving along a line with the given velocity v. Assume t is time measured in seconds and velocities have units
Lady_Fox [76]

Answer:

a. the motion is positive in the time intervals: [0,2)U(6,\infty)

   The motion is negative in the time interval: (2,6)

b. S=7 m

c. distance=71m

Step-by-step explanation:

a. In order to solve part a. of this problem, we must start by determining when the velocity will be positive and when it will be negative. We can do so by setting the velocity equation equal to zero and then testing it for the possible intervals:

3t^{2}-24t+36=0

so let's solve this for t:

3(t^{2}-8t+12)=0

t^{2}+8t+12=0

and now we factor it again:

(t-6)(t-2)=0

so we get the following answers:

t=6  and t=2

so now we can build our possible intervals:

[0,2)  (2,6)  (6,\infty)

and now we test each of the intervals on the given velocity equation, we do this by finding test values we can use to see how the velocity behaves in the whole interval:

[0,2) test value t=1

so:

v(1)=3(1)^{2}-24(1)+36

v(1)=15 m/s

we got a positive value so the object moves in the positive direction.

(2,6) test value t=3

so:

v(1)=3(3)^{2}-24(3)+36

v(3)=-9 m/s

we got a negative value so the object moves in the negative direction.

(6,\infty) test value t=7

so:

v(1)=3(7)^{2}-24(7)+36

v(1)=15 m/s

we got a positive value so the object moves in the positive direction.

the motion is positive in the time intervals: [0,2)U(6,\infty)

   The motion is negative in the time interval: (2,6)

b) in order to solve part b, we need to take the integral of the velocity function in the given interval, so we get:

s(t)=\int\limits^7_0 {(3t^{2}-24t+36)} \, dt

so we get:

s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{7}_{0}

which simplifies to:

s(t)=[t^{3}-12t^{2}+36t]^{7}_{0}

so now we evaluate the integral:

s=7^{3}-12(7)^{2}+36(7)-(0^{3}-12(0)^{2}+36(0))

s=7 m

for part c, we need to evaluate the integral for each of the given intervals and add their magnitudes:

[0,2)

s(t)=\int\limits^2_0 {(3t^{2}-24t+36)} \, dt

so we get:

s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{2}_{0}

which simplifies to:

s(t)=[t^{3}-12t^{2}+36t]^{2}_{0}

so now we evaluate the integral:

s=2^{3}-12(2)^{2}+36(2)-(0^{3}-12(0)^{2}+36(0))

s=32 m

(2,6)

s(t)=\int\limits^6_2 {(3t^{2}-24t+36)} \, dt

so we get:

s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{6}_{2}

which simplifies to:

s(t)=[t^{3}-12t^{2}+36t]^{6}_{2}

so now we evaluate the integral:

s=6^{3}-12(6)^{2}+36(6)-(2^{3}-12(2)^{2}+36(2))

s=-32 m

(6,7)

s(t)=\int\limits^7_6 {(3t^{2}-24t+36)} \, dt

so we get:

s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{7}_{6}

which simplifies to:

s(t)=[t^{3}-12t^{2}+36t]^{7}_{6}

so now we evaluate the integral:

s=7^{3}-12(7)^{2}+36(7)-(6^{3}-12(6)^{2}+36(6))

s=7 m

and we now add all the magnitudes:

Distance=32+32+7=71m

7 0
3 years ago
There are 18 pencils for $2.36 how much would it be for one
stira [4]

Answer:

13 cents

Step-by-step explanation:

2.36/18= 0.13

5 0
3 years ago
Which of the following number sentences illustrates the associative property of multiplication?
Mila [183]
The correct answer is c :)
5 0
3 years ago
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